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These are replies submitted by Hullzie16


I omitted all of the physics which looking back might have been a bad idea. 

So what I am interested in doing is solving for the metric function f(r) which is a solution to the modified graviational field equations which are R[a,b]=0. But then again the field equations could also be written as R[~a,b]=0 or even RR[~a,~b]=0. So the issue is it does not matter how you write the equations when you are trying to solve for the functions. 

I am not suggesting that R[1,1]=R[~1,1]. I am saying that the solution for f(r) from the field equations should be the same regardless of how we solve - they are not. Secondly if we use the definition R[~1,1] = g_[~1,~a]R[a,1] You can see that this does not match what we get from the tensor I define. I did this in a modified sheet which no longer has the cross-term in the metric so we avoid the taylor series.

Can you clarify some of your notation. Is u an unkown function, does it depend on all the variables? What does u_x represent? 

Do you want L to operate on T and vice versa or is it multiplication after it operators on some function? 

If you can clarify this I think I can write something down relativietly quick, in the mean time you can always do something like this. 



F(x)*`will now be displayed as`*F



x, y, z



u^2*%diff(F(x, y, z), x, x)+%diff(F(x, y, z), y)






To create operators. 


Thank you both @acer @Rouben Rostamian for your replies. 

I will repost. 


In your second problem you have not specified n, S and Nc. If you do that from what you say in your "a" defintions it runs for me. 

Always make sure there are no unkown variables in your boundary conditions otherwise it will not run. 

See attached.

If you can upload your worksheet with your attempted solution, or provide more details to what you exactly are looking to solve this would be helpful. 

The big green arrow is used for uploading worksheets or the contents of your sheet. 


You have some issues in your physics. The final solution in your image is not the solution of a spacetime with that stress energy distribution. That solution solves the vacuum field equations with a cosmological constant. 

Please keep on eye on this thread as I will write an updated solution shortly. 


Thanks for reminding me of that option, I fail to remember that option exists when I do calculations with Lambda. 

Can you give some more information about the stress-energy tensor? It is incomplete currently but I believe you are using a perfect fluid? 

If you could also upload your attempts that would be great. 

Lastly here is the method for solving just the 2+1 with a cosmological constant and no stress-energy. Once you give more information of confirm it is a perfect fluid I will update the response.


The green arrow that appears at the top of the text box (see image below) 

Click on it, go to choose files as it says and click on your saved worksheet then click upload and insert link. 

You will get the best responses if you post your worksheet to show what you have attempted. 

Or at the vary least post the ODEs you are working with. 

You can use the green arrow to upload your worksheet or images. 


Hello, I am sorry for replying so late. 

The desired output should be:

Details can be found in: where some terms have been ignored. 

Thank you for taking a look into this. 


Just out of curiousity what are the exact ODEs you are trying to solve? 


It does do rather interesting things in the negative region..something I had not thought about.

Is there any portion of the code that does deal with stepsize/speed? Not that I want to "speed up" the process just more of a curiousity on how it works. 

The positive result is also interesting, not what I expected. I will keep tinkering with the parameter ranges and see if anything can occur. I appreciate this worksheet, it appears my attempt was far too simplistic. 



As mentioned in my comment above I am still have issues with CW suggestion and still recieving errors. This is a much more involed solution then I was expecting although much more elegant than my attempts. I have some questions however, some may be minor in nature. 

This method replaces my looping attempt by plotting the region of which the inequlaity you is satisfied in blue by examining the region alpha=-1.5..0 and beta -2..0 in step sizes of epsilon =e^-5? Is that the correct understanding?

This reason I ask if because I am interested in regions of positive alpha and beta only and switching to alpha=0..1.5 and beta=0..2 and running it has yet to produce a result after approximately 15 minutes real time. So if i wanted to "speed up" could I make the epsilon larger at the cost of precision? 


@Christian Wolinski 

Using this substitution I still recieve an error - for some reason I cannot upload worksheet contents only the worksheet link itself. It appears to not be able to actually compute the integration as I recieve the new error in the modified sheet below.

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